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antiseptic1488 [7]
2 years ago
15

Scientists developed a device capable of lowering temperatures in order to slow molecular motion to almost a complete stop. Whic

h of these temperatures is theoretically able to make this happen?
0 K

0°C

273 K

273°C
Chemistry
1 answer:
seropon [69]2 years ago
3 0

Answer:

273k

Explanation:

Soooo... This is a complicated problem.

I dunno. Like... Is this a trick question? Trivia, maybe?

Nahhhhh...

It's just not telling us facts...SOME facts, but we need to use past learning to help us.

FIRST of all, what the heck does K even mean?

Is this a memo? A demo? A KO?

Okay, it's nither of these things.

So, let me tell you what the "K" means.

The K means "Kelvin" which is basically a word form for "-"

I AM NOT MAKING A FACE. IT'S THE WORD FORM FOR MINUS. STOP LAUGHING!!!!

And so... To bring objects to a COMPLEETE SKRRRRRRRRRRRRRRRRRRRRRRT...

It will need to be -273.

This means that the answer is 273k.

Which reminds me, Ima play 2k.

And ima rap to-day.

And now I'm being lame.

Just answer the question :/

You might be interested in
ASAP , 8.81 g Carbon
xz_007 [3.2K]

The empirical formula : C₂Cl₇

The molecular formula : C₁₀Cl₃₅

<h3>Further explanation</h3>

Given

8.81 g Carbon

91.2 g Chlorine

Molar Mass: 1362.5 g/mol

Required

The empirical formula and molecular formula

Solution

Mol ratio :

C = 8.81 g : 12.011 g/mol =0.733

Cl = 91.2 g : 35,453 g/mol = 2..572

Divide by 0.733

C : Cl = 1 : 3.5 = 2 : 7

The empirical formula : C₂Cl₇

(The empirical formula)n = the molecular formula

(C₂Cl₇)n = 1362.5

(2x12.011+7x35.453)n=1362.5

(272.193)n=1362.5

n = 5

6 0
3 years ago
Now consider the reaction a+2b⇌c for which in the initial mixture qc=[c][a][b]2=387 is the reaction at equilibrium? if not, in w
Paul [167]
Solving this chemistry is a little bit hard because the question didn't give some important detailed. 
So first, there are a couple problems with your question. 
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be: 
Kc = [C] / [A] [B]^2 
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.
5 0
3 years ago
Part A The first step to engineering is to define the problem. Write down the problem the students have to solve, and describe t
Elena L [17]

Answer:

Engineering is all about solving problems using math, science, and technical knowledge. And engineers have solved a lot of problems in the world by designing and building various technologies. We have everything from machines that can breathe for you in hospitals to suspension bridges to computers we use every day. All of these things were once designed by engineers using the engineering design process.

Explanation:

3 0
2 years ago
Read 2 more answers
Which substance will most likely dissociate when it is dissolved in water?
Katyanochek1 [597]

Answer: CaC12

Explanation: Calcium chloride not 100% sure tho

6 0
3 years ago
Read 2 more answers
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
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